# Solve for x 2x/(x+5)+ 10/(x^2+2x-15)=x/(x-3)

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When we try to add or subtract fractions, all ratios have to have the same denominator.

The denominator of the ratio 10/(x^2+2x-15) is a quadratic.

We'll check if the quadratic has real roots. If it has, we can re-write it as a product of linear factors.

To verify if it has real roots, we'll apply the quadratic formula:

x1 = [-b+sqrt(b^2 - 4ac)]/2a

2x/(x+5) + 10/(x^2+2x-15) = x/(x - 3)

We'll identify a,b,c:

a=1

b=2

c=-15

x1 = [-2+sqrt(4+60)]/2

x1 = (-2+8)/2

x1 = 3

x2 = -5

We'll re-write the quadratic:

x^2+2x-15 = (x-3)(x+5)

We'll re-write the expression:

2x/(x+5) + 10/(x-3)(x+5) = x/(x - 3)

In order to find out x, we'll multiply the first ratio by (x-3) and the third ratio by (x+5):

2x(x-3) + 10 = x(x+5)

We'll remove the brackets:

2x^2 - 6x + 10 = x^2 + 5x

We'll subtract x^2 + 5x both sides:

x^2 - 11x + 10 = 0

We'll apply the quadratic formula, again:

x1 = [11+sqrt(121-40)]/2

x1 = [11+sqrt81]/2

x1 = (11+9)/2

x1 = 9

x2 = (11-9)/2

x2 = 1

The real roots of the given expression are: {1 ; 9}